First person to solve this for me will receive 0.004 BTC

t = r*F*sin(x)
x = 45deg
sin(45deg) = sqrt(2)/2
r = 280cm
F = 1000kg*g = 9807N
t = 19417 Nm

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There is a weight of 1000kg 40cm behind the rear wheel that is laterally equally distributed... No such data has been given about the weight of the car.... More specifically the front end weight

You need to learn how to read, my friend.
Both of these are almost definitely definitely wrong.

((875)kg*(0.160^2)m)/(2700^2)s = 3.07Nm

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The rear moment counteracts the moments from the first 40 centimeters. We only need to consider the weight from 80cm to 320cms from the wheel. These moments average out to a moment at 200cm of 750 kg. This is a moment of 150Nm.
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The 45 degrees is irrelevant to the question because the car initially must life from 0 degrees. My initial answer was wrong, forgot gravity. 1470 N-m

19wkoLCbVMLBmBTm9CGyKWSmKNgDwzk2uM

If the car needs to sustain 45 degrees it only needs to produce sin(45)*1470.

This is impossible, you don't know the weight of the driver, fuel load, you don't know how much grip the tyres are capable of producing,

>almost definitely

whoops you seem to have typed some retardation, did you want help fixing that?

Do your own homework

12,26 kNm to lift off
8671 Nm to remain up

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I presume the exercise only demands the torque needed for the car to stay upright at 45 degrees.

The answer is 16970.55Nm

The exis of rotation, judging from the instructions, is the rear axle.

Thus, we have a = 280 cm and m1 = 875 kg in from of the axle and b = 40 cm with m2 = 125 kg behind it.

So, when the car is upright at fi = 45 degrees, there are two torques: the one in front and the one behind.

In front: M1 = m1 * g * a * sin(fi) = 17 324.1 Nm
Behind: M2 = m2 * g * b * sin(fi) = 353.55 Nm

Now these torques have the opposite direction. The net torque is then M1 - M2 = 16970.55 Nm, pulling the front down. You need the same amount at the rear axle to keep it at the angle.

what kind of tires is OP using? if normal profile tires no amount of torque is going to do shit for you, you will just do a burnout every single time and not a wheelie

All answers here are wrong, you are all doimg calculations by asumim that car is 2 particles that are some distance from axis. To do this calculation you need to calculate moment of inertia for thin rod (that must be assumed since op didnt say anithing about height or with) and then you ballance the torques.

he's right I forgot they're opposing so:
takeoff: 11,8 kNm
stay at 45°: 8324 Nm
(you forgot that you have to divide the distance by 2 because the weight is in the middle)
I'm still the first with the correct solution

also mfw OP is posting this from his FWD Nissan Cuckwagon dreaming about an actual car

its a physics question, not a real world one..
OP wants to know what torque a car needs to do this. so there are 2 solutions here: the amount it needs to get moving, and the amount it needs to stay up. get your dynamics out of here.

Weight is uniformly distributed, its not in the middle.

well you're saying that all weight is in the bumpers, that doesn't sound like uniform distribution to me.

have you ever attended a physics class?

>its a physics question, not a real world one..
this is a stupid fucking thread and on a wrong board

What i am saying it that to get the rigt answer you must use formula T=Ia where I is moment of inertia of a thin rod, noone here seams to underatand basic solid body physics. You cant just say T=La for solid object, this only works for particles. Fuck i the reward was bigger i would have solved this problem but watchig you all strugle is the reward in itself

Why dont you set the problem infitesimaly and do the integral, then come back and se if those answers are right.

Damn it. Where is your BTC address though?

"how much needs to be produced" was the question. it's not about dynamics.the point is to calculate the current torque due to gravity, which gives 11,8 kNm. any torque higher than that will lift the nose up. it WILL move. you just have get the throttle down to stop at the right point. great that you know about inertia, but it is not required here.
not expecting OP to deliver honestly
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Right here OP

0xe4e6bF2EF8ddE56e077C115d219f75F4FA0cE1db

isn't that an ETH address? also, there are IDs here you know

This guy gets it...

Anyone attempting to answer this question with unsubstantial data being provided should just GTFO

The only equation that needs to be solved is how to get my dick near some pussy

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